Question

A cistern has three pipes A,B,C. A andB can fill it in 3hours and 4 hours respectively while C can empty the completely filled cistern in 1 hour. If the pipes are opened in order at 3 PM , 4PM and 5PM. respectively, at what time will the cistern be empty

Answer 1

This is the answer of this question

Answer 2

The tank will be empty at 7:12 pm

Pipe A can fill in one hour = 1/3 = 4/12

Pipe B  can fill in one hour = 1/4 = 3/12

Since, pipe B is opened at 4pm, then till 5pm, the total will be -

= A+B = 4/12+3/12

= 7/12 of the work(to fill)

Thus, the total work done till 5 pm is

= 4/12+7/12 = 11/12

Pipe C is opened at 5pm and empties in one hour, thus

= -12/12

Work done by A+B+C after 5 pm-

= 4/12 + 3/12 - 12/12

= -5/12

Work done at 6pm -

= 11/12 - 5/12=

6/12

= 1/2

Work done at 7pm,

= 6/12 - 5/12

= 1/12

Thus, in one hour, -5/12 work is being done. In each 12 mins, -1/12 work is done.

Therefore, the tank will be empty at 7:12 pm